Standard Deviation

Date: 07/05/97 at 05:49:21
From: Jeremy Bowell
Subject: Averages involving standard deviation
How do I solve this problem?
The average height of year 10 students is given as 175 cm and the
standard deviation is 12cm. Find the percentage of students whose
height is:
a) Greater than 175 cm
b) Between 163 and 187 cm
c) Greater than 187 cm
d) Between 151 and 163 cm
What is standard deviation?

Date: 07/05/97 at 07:58:37
From: Doctor Anthony
Subject: Re: Averages involving standard deviation
Standard deviation is a measure of the spread of a distribution about
the mean. It is the square root of the VARIANCE, where variance is
calculated from:
VAR(x) = E(x^2) - Mean^2
E(x^2) is the 'expected value' of x^2. If this term is unfamiliar to
you, consult a standard textbook or write back.
(a) Because the normal distribution is symmetric about the mean, the
percentage of students with height greater than the mean, 175 cm, is
50 percent.
Using the normal tables, let m = mean and s = standard deviation. Z
values are those on the horizontal axis giving the number of standard
deviations from the mean, with z = 0 at the mean. Areas are the areas
under the normal probability curve between two z values. These areas
are found using normal tables and entering the appropriate z values.
(b) You calculate the z values using:
x - m 163 - 175
z = ------ = --------- = -1
s 12
The area between the mean and -1 s.d. = .3413
187 - 175
Also for the upper limit z = ----------- = +1
12
Again, the area between the mean and 1 s.d. = .3413
Total probability is then 2 x .3413 = 0.6826, which is 68.26 percent.
(c) Greater than 187 gives area 0.5 - .3413 = .1587, so the number
greater than 187 is 15.87 percent.
(d)
151 - 175
z = --------- = -2 area between mean and -2 is 0.47725
12
163 - 175
z = ---------- = -1 area between mean and -1 is 0.3413
12
The area between -2 and -1 is then .47725 - .3413 = 0.13595
So 13.6 percent of students have heights between 151 and 163 cms.
-Doctor Anthony, The Math Forum
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